# Volts x Resistance = A Thing?

And yeay, U is what we Europeans use.

Hey, hey!, I am spaniard, and we used "V" from elementary school to the grade. ;)

Regards

Do you also use a zigzag for a resistor? ;)

Do you also use a zigzag for a resistor?

Indistinctly: a zig zag or a rectangle. (I am starting to worry about my technical habits . . . . . . maybe the american influence . . . .)

It used to be a zigzag but they changes it I think in the 70s to a rectangle. I still do a zig zag for hand drawn circuits but a rectangle when using a CAD package.

Pin the UK we don't use U for voltage but V, however sometimes it is E for electro motive force, so we would say ohms law I'd E = I * R The reason A is not used is that it was already taken in Physics for acceleration.

For extra credit, show that the units on both sides of E = mc2 are equal.

Considering that “c” = the speed of light (299 792 458 m / s), and

c2= 89875517873681764,

E, is going to be very large no matter what “m” (mass) is.

Back to you Chet…

What is 2250? is it just a thing, or is it something? I’m guessing some kind of resistance?

If you look at the voltage divider, the base resistor (the one connected to ground), divided by the
total resistance, gives the ratio that represents by what amount the input voltage has been divided.
The “2250” you inquire about is , by itself, nothing, because the division should be performed FIRST. The input voltage (5V) should NOT be multiplied by the base resistor first as you have done.
You should calculate the divider ratio [R2/(R1+R2)] , BEFORE multiplying it by the input voltage (5). Why ? because the objective is to determine the divider ratio so you can test different input voltages (5,10 ,25, 20 etc…) or (5,4,3,2,1 etc).

Funny, all the European (Electrical Engineering) books I have talk about U and I don't think they're all from Dutch authors (all written in English of course). But yeay, try talking to a physics engineer or mathematical engineer about imaginary numbers :D

raschemmel:
Considering that “c” = the speed of light (299 792 458 m / s), and

c2= 89875517873681764,

E, is going to be very large no matter what “m” (mass) is.

Unless it’s a really small mass. :-)

Doesn’t it start out with the first atom and then continue on in a “chain reaction” to the rest of the weapons grade material used in “the device” (or was it “the gadget”) ? (meaning “m” is actually natoms*mof material used(ie: plutoneum))

Vout = (Vin x R2) / (R1 + R2)

I think the parentheses around Vin x R2 are causing some confusion.

R2/(R1+R2) is a ratio of R2 to the total resistance, multiplied by the applied voltage to get the unloaded voltage on R2. So conceptually it would be: Vin x (R2/(R1+R2))

Or you may look at it as the current through the resistors, Itot = Vin/(R1+R2), multiplied by R2 to get the voltage drop across R2. R2 x (Vin/(R1+R2))

raschemmel:
Doesn’t it start out with the first atom and then continue on in a “chain reaction” to the rest of the weapons grade material used in “the device” ? (meaning “m” is actually natoms*mof material used(ie: plutoneum))

E = mc2 is more general than a fission reaction.

A single electron and positron annihilation releases net energy equivalent to their rest masses, something around 1 Mev. You’d need on the order of 1018 of those just to heat up a cup of coffee.

raschemmel:
Doesn’t it start out with the first atom and then continue on in a “chain reaction” to the rest of the weapons grade material used in “the device” ? (meaning “m” is actually natoms*mof material used(ie: plutoneum))

No not quite. It depends on if you are dealing with a fission bomb or fusion bomb but a whole atom is never converted into energy. The energy comes from the fact that if you fuse four hydrogen atoms into one helium atom, that atom is slightly lighter than the total mass of the four helium atoms that went into it. It is this difference in mass that provides the m in mc2.
Oddly enough at the other end the same thing happens, split a heavy atom up and the two pieces ( atoms ) weigh less than the original atom.
This stops at iron which is the lowest energy matter state so fission of iron requires energy to make it happen as does fusion of iron. That is why there is such a lot of it in the universe.

septillion: Funny, all the European (Electrical Engineering) books I have talk about U and I don't think they're all from Dutch authors (all written in English of course). But yeay, try talking to a physics engineer or mathematical engineer about imaginary numbers :D

Well never seen it called U in England.

As to imaginary number Electronic engineers couldn't use I because that was already nabbed for current, both Physicists and Mathematicians call it I and Electronic Engineers call it J

Yeah, make that electronic engineers call it j, the rest of the world calls it i :p (And to be picky, not capital. But now I'm just teasing.)

Grumpy_Mike: This stops at iron which is the lowest energy matter state so fission of iron requires energy to make it happen as does fusion of iron. That is why there is such a lot of it in the universe.

It's actually an isotope of nickel that has that distinction. The greater abundance of iron requires further explanation.

Complex Numbers

@ jboyton, I have a question you could probably answer. At work we are replacing a zinc plated wire terminal with a nickle plated terminal (to avoid rust and corrosion) I did a test with the zinc plated terminal immersed in water for a week and it started to oxidize. I told the Mechanical Engineer I needed to redo the test with Arrowhead drinking water because our concern was related to rain , not tap water and the arrowhead was probably closer to rain. He said the rain was probably more contaminated that the tap water . Any thoughts ?

I really don’t know. I’d guess that rainwater is more contaminated. But I’ll bet it’s quite variable depending on where you collect it and how you collect it and what sort of contaminants you care about. By the same token “Arrowhead drinking water” probably isn’t exactly the same thing from place to place.

I’m guessing (again) that pH is important when it comes to corrosion. Maybe it makes sense to measure some of the actual rainwater? Just a thought from someone who knows nothing about water chemistry.

jboyton: It's actually an isotope of nickel that has that distinction. The greater abundance of iron requires further explanation.

It is generally believed that iron-56 is more common than nickel isotopes in the universe for mechanistic reasons, because its unstable progenitor nickel-56 is copiously made by staged build-up of 14 helium nuclei inside supernovas, where it has no time to decay to iron before being released into the interstellar medium in a matter of a few minutes, as the supernova explodes. However, nickel-56 then decays to cobalt-56 within a few weeks, then this radioisotope finally decays to iron-56 with a half life of about 77.3 days. The radioactive decay-powered light curve of such a process has been observed to happen in type II supernovae, such as SN 1987A. In a star, there are no good ways to create nickel-62 by alpha-addition processes, or else there would presumably be more of this highly stable nuclide in the universe.

I knew thaat...

It is generally believed that iron-56 is more common than nickel isotopes in the universe for mechanistic reasons, because its unstable progenitor nickel-56 is copiously made by staged build-up of 14 helium nuclei inside supernovas, where it has no time to decay to iron before being released into the interstellar medium in a matter of a few minutes, as the supernova explodes. However, nickel-56 then decays to cobalt-56 within a few weeks, then this radioisotope finally decays to iron-56 with a half life of about 77.3 days. The radioactive decay-powered light curve of such a process has been observed to happen in type II supernovae, such as SN 1987A. In a star, there are no good ways to create nickel-62 by alpha-addition processes, or else there would presumably be more of this highly stable nuclide in the universe.

The OP asked of volts being multiplied by ohms . . . . .what a great forum is this!

Yes we did that at the very beginning :)